Nuclear and Particle Physics
1
Nuclear Physics
Back to Rutherford and his discovery of the
nucleus
Also coined the term “proton” in 1920, and
described a “neutron” in 1921
Neutron discovered by Chadwick in 1932
Ernest
Rutherford
1871-1937
me = 9.1 x 10-31 kg
mN = 1.6749 x 10-27 kg
mP = 1.6726 x
10-27 kg
nucleons
James Chadwick
1891-1974
2
Nuclides and Isotopes
To specify a nuclide:
A
Z
X
Z is the atomic number = number of electrons or protons
A is the mass number = number of neutrons + protons
So number of neutrons = A-Z
Number of protons = Z
Isotopes – same atomic number, different mass number
e.g. carbon:
12
6
C
13
6
C
Many isotopes do not occur naturally, also elements > U
3
Sizes
We saw with the Bohr model
that radius of the atom
depended on atomic number
Nucleus = protons + neutrons
= mass number
The volume of a nucleus is
proportional to the mass
number
V

15
r  (1.2 x10 m) A
3
4 3 4
π r  π (1.2x10 15 )3 A (m3 )
3
3
4
Masses
Mass spectrometer
1 atomic mass unit (u.) = 1.6606 x 10-27 kg = 931.5 MeV
Fixed so that carbon = 12.00000 u
mN = 1.6749 x 10-27 kg = 1.0087 u
mP = 1.6726 x 10-27 kg = 1.0078 u
5
Binding Energy
Total mass of a nucleus < sum of masses
Example:
Mass of helium nucleus = 6.6447 x 10-27 kg
4
2
Contains 2 protons and 2 neutrons
He
Mass = 2 x (1.6749 x 10-27 + 1.6726 x 10-27 ) kg
= 6.6950 x 10-27 kg
Difference = (6.6950 – 6.6447) x 10-27 = 0.0503 x 10-27 kg
Energy = mc2 = 0.0503 x 10-27 x c2
= 4.53 x 10-12 J
= (4.53 x 10-12) / (1.6 x 10-19) = 2.83 x 107 eV
= 28.3 MeV
6
Atomic Mass Units
1 u = 931.5 MeV
mN = 1.6749 x 10-27 kg = 1.0087 u
mP = 1.6726 x 10-27 kg = 1.0078 u
Mass of helium nucleus = 4.0026 u
7
Atomic Mass Units
Same calculation
Mass of 2p + 2n = 2 x (1.0078 + 1.0087) = 4.0330 u
4.0330 u
Difference = 0.0304 u
Binding energy = 0.0305 x 931.5 = 28.3 MeV
8
Average Binding Energy
He – 4 nucleons, 28.3 MeV total: average = 7.075MeV
• Graph
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Attractive?
How does nucleus stay together?
Like charges repel!
Force stronger than electric force
Strong nuclear force
Short range (~10-15 m)
Stable nuclides N = Z
A > 30-40 – more neutrons
Z > 82 – no stable nuclides
Strong force can’t overcome repulsion
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Radioactivity
Becquerel, 1896
Emission of radiation without external stimulus
Curies – polonium (Po) and radium (Ra)
Radioactivity unaffected by heating, cooling,
etc.
Henri
Becquerel
1852-1908
1903 (Physics)
1911 (Chem)
Marie Curie
1867 - 1934
Pierre Curie
1859 - 1906
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Classification
Rutherford classified 3 types of radioactivity
according to penetration power
Also different charge
Video:
“People Pretending to
be Alpha Particles”
Important factor: Conservation of nucleon number
(neutrons + protons) = (neutrons + protons)
12
Alpha Decay
Least penetrating – nucleus of
4
2
He
Radium 226 is an alpha emitter:
226
88
Parent
Ra 
222
86
Rn  He
4
2
transmutation
Daughter
Mass of parent > mass of daughter + mass of alpha
Difference = kinetic energy
13
Example
232
92
232.03714 u
U
Th  He
228
90
4
2
 228.02873 u + 4.002603 u
total = 232.03133 u
Lost mass = 232.03714 – 232.03133 = 0.00581 u
0.00581u x 931.5 MeV/u = 5.4 MeV
(some recoil)
14
Beta decay
 One electron
14
6
C N
14
7
0
1
e
What is lost is NOT an orbital electron
Instead a neutron changes to a proton + electron
So (6p + 8n) => (7p + 7n) + e-
-
decay
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Example
14
6
C N
14
7
0
1
e
Keep track of electrons!
Carbon 14 has m = 14.003242 u 6 electrons
Nitrogen 14 has m = 14.003074 u normally 7 electrons
But in the decay, the nitrogen would have 6 electrons
However the total on the r.h.s. of the equation has 7
So difference = 0.000168 u = 0.156 MeV = 156 keV
16
Conservation of energy
• Energy of decay = 156 keV = problem!
?
17
A new particle
•
•
•
•
Proposed by Pauli (1930) - neutrino
Theory by Fermi
Discovered 1956
Zero charge, ~0 rest mass
14
6
C N
14
7

0
1
e
Wolfgang Pauli
1900-1958
antineutrino
Cosmic neutrino
detection
“Zero rest mass” – speed of light
1998 – Super Kamiokande – some mass
Enrico Fermi
1901-1954
18
More on positrons
Many isotopes have more neutrons than protons
 Decay by emission of electron
Other isotopes have more protons than neutrons
 Decay by emission of positron
Proton changes to a neutron + positron
19
10
Ne  F  e  
19
9
0
1
+ decay
19
Annihilation
Proton changes to a neutron + positron
19
10
Ne  F  e  
19
9
0
1
+ decay
Positron annihilation
Application – positron emission tomography
20
Positron Emission Tomography
PET – basis – use radio-labelled compounds, i.e. those
containing a radionuclide.
Positron emitters:
15
8
O,
13
7
11
6
N, C,
18
9
F,
124
53
I
As an example, oxygen-15 can be used to
look at oxygen metabolism and blood flow.
Fluorine-18 is commonly used to examine
cancerous tumours.
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PET - method
Annihilation produces two back-to-back 511 keV photons
Simultaneous detection
22
Electron capture
• Nucleus absorbs orbiting electron
7
4
Be 
0
1
e  Li  
7
3
Proton changes to neutron
Usually K electron
X-ray emission as outer electron
jumps down to K
23
Gamma decay
Most penetrating
 = photon. High energy
*Excited nucleus  lower energy state
12
5
B C
12
5
B  C* 
12
6
0
1
12
6

12
6
e
0
1
- (9.0 MeV)
e
- (13.4 MeV)
C
 (4.4 MeV)
Energy levels far apart = keV or MeV
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Homework . . .
1. p.902,#6;
2. p.908, Practice 25B;
3. p.912,Section Review
4. p.928, 30-37;
5. p. 930, 56,60;
6. Read through lab for next
time; answer pre-lab
questions
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